Triangular inequality extended Let $a,b,c,d,e,f\in\mathbb{R}$ all positive or zero, such that $a\leq c+d$ and $b\leq e+f$ show that: $$\sqrt{a^2+b^2}\leq\sqrt{c^2+e^2}+\sqrt{d^2+f^2}$$ Some hint or auxiliar inequality that would help? I've done many attempts but write them will take a lot of time, I tried to use the artihmetic and geometric mean, minkowski but no one works (or fits) can you help me?

Just $$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}\geq\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$ If I don't see the triangle inequality I can make the following.

By C-S $$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}=\sqrt{c^2+e^2+d^2+f^2+2\sqrt{(c^2+e^2)(d^2+f^2)}}\geq$$ $$\geq\sqrt{c^2+e^2+d^2+f^2+2(cd+ef)}=\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$